# Homework Help: Triangle problem

1. Dec 23, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
A point P is located in the interior of an equilateral triangle ABC. Perpendiculars drawn from P meet each of the sides in points D,E,F, respectively. Where should P be located to make PD + PE + PF a maximum? What about a minimum?

2. Relevant equations

3. The attempt at a solution

From symmetry considerations. I think that the maximum will be at the incenter. I am not sure about the minimum. Maybe at a vertex, but that is not really in the interior... I tried the standard reflections. I am not sure if I actually need to set up a coordinate system with P at the origin... That looks like it could get messy? I am also not sure how to use the fact that this is an equilateral triangle.

2. Dec 23, 2007

### ehrenfest

anyone?

3. Dec 24, 2007

### marlon

Isn't this one of them Lagrange multiplier things ?

Having said that, i don't know how to set up :

1)the funtion that needs to be minimized/maximized

2) the function that is going to be the constraint.

I would reallly like to see the solution to this problem, though.

Marlon

4. Dec 24, 2007

### jacobrhcp

I think it's my english that's bad here, but how can you draw 3 perpendicular lines in a plane?

5. Dec 24, 2007

### chickendude

They are not perpendicular to each other.

They are perpendicular to each side of the triangle.

6. Dec 24, 2007

### chickendude

How about this approach: (sorry for the double post)

Draw segments from P to each vertex of the triangle. You have split the triangle into three new triangles.

The area of each triangle is given by $$\frac{1}{2}sh_i$$ where s is the side length of the equilateral triangle and h_i is the height of the perpendicular on that side

Through this, we can set up the equality

$$\frac{1}{2}sh_a + \frac{1}{2}sh_b + \frac{1}{2}sh_c = \frac{s^2 \sqrt{3}}{4}$$

$$h_a + h_b + h_c = \frac{s \sqrt{3}}{2}$$

meaning that the sum of the three perpendiculars is always constant regardless of the location of P

7. Dec 24, 2007

### jacobrhcp

well, take the situation that P is at point a, then if you move D a distance d, for instance, you can still hold E still, but you'll have to move F to keep P one point.

now P moves in the direction of E with a distance: 2d/root[3], this is the distance it shrinks.
the distance it grows is: 2d tan(30) = 2d/root[3], this is equal.

because any motion of P can be described as the sum of the motion in the direction of perpendiculars, the distance is always the same.

Last edited: Dec 24, 2007
8. Dec 24, 2007

### jacobrhcp

I must have done something wrong, because chickendude seems right.

EDIT: I've edited my post and saw what i did wrong.

Last edited: Dec 24, 2007
9. Dec 24, 2007

### Rainbow Child

10. Dec 24, 2007

### ehrenfest

Hmmm. I did not see anything wrong with chickendude's post, which means there is probably something wrong here.

11. Dec 24, 2007

### NateTG

If you consider the point $p$ to be the tip of three triangles, each of which has one of the sides of the original equilateral triangles as it's base then the total area of those triangles will be:
$$\frac{1}{2} l_1 s + \frac{1}{2} l_2 s + \frac{1}{2} l_3 s = \frac{1}{2} (l_1 + l_2 +l_3) s$$
which is also the area of the original triangle, which is clearly constant, so
$$(l_1+l_2+l_3)$$
must be constant.

12. Dec 24, 2007

### ehrenfest

Yes. You just repeated chickendude's proof. I am just saying if that is right, then something is wrong with Rainbow Child's proof.

13. Dec 24, 2007

### Rainbow Child

My mistake!

In my proof I took the points

$$E(\frac{\alpha}{4},\frac{\sqrt{3}\,\alpha}{4}),\,F (\frac{3\,\alpha}{4},\frac{\sqrt{3}\,\alpha}{4}),\ ,D(\frac{\alpha}{2},0)$$

which are the midpoints of the sides AB, AC and AD! (nothing to do with the original problem )

So I gave the answer to the question:

"Which point extemizes the sum $$|PD|+|PE|+|PF|$$ when D, E and F are the midpoints of the sides!"

Sorry!

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